# Modified Gummel-Poon BJT Modelling Issue

First of all, sorry for my bad english.

Output for Vb:

From the schematic above, I redrew the 2N3904 BJT into the modified Gummel-Poon model.

Output for Vb:

The equation for B1, B2 and B3 were determined using the info below.

Then, the values of the intrinsic capacitance were determined using the DC operating point analysis in LTSpice.

Cbc = C2

Cbe = C3

But the result seems wrong for me. What do I miss? Can I just take and use the values of Cbc and Cbe from the DC operating analysis log file of LTspice?

• If you're going to use behavioural sources with quantities involving nodes from the schematic (e.g. V(n005)), then it is (almost) mandatory you manually rename the nodes. This is because, as you add more nodes, their names are automatically re-ordered, so what you think as V(n005), it may no longer be the case after any operation involving a modification of the nodes in the schematic. It also makes debugging easier, not to mention it makes it more clear for people reading your schematic which nodes you're referring to. Apr 2, 2020 at 16:12
• Thanks for replying. I did check the node number and equation before simulating my circuit. The steady-state values of base voltages in both schematics are actually identical. It's just that the time taken to reach the steady state are different. Hence, I guessed the problem is caused by the wrong capacitance value. Capacitance values do not have effect when it is in steady state (frequency = 0). I am wondering what is the correct way to determine the parasitic capacitance. Is my step in determining parasitic capacitance correct? Apr 3, 2020 at 2:22

If it's SPICE world then this setup may be a bit more helpful: a current source directly in the base of the transistor, with either the emitter (for determining $$\C_{je}\$$), or the collector (for $$\C_{jc}\$$) grounded, and the results come by reading the voltage. Or a voltage source and read the current. Here's an example with 2N2222, with the emitter grounded:
You know that $$\R_b\$$ is in series with $$\R_e\$$ and $$\C_{je}\$$. This is a series RC which, when fed with a constant current source, becomes a PI compensator, so you know you have a corner frequency given by $$\f_c = (R_b + R_e)C_{je}\$$. Until $$\f_c\$$ it's an integrator response, afterwards it's flat. The flat portion tells you the DC voltage given by $$\R_b + R_e = 10.2\textrm{V}\$$ (the 1st cursor), and $$\f_c \approx 624\textrm{MHz}\$$ is given by reading the phase of $$\-45^\textrm{o}\$$ (1st order RC, cursor 2, Y-axis set to linear, for easier reading). Since the current source uses AC 1, that means that the input is unity, so there's no need to divide the voltage by the input current.
$$C_{je}=\frac{1}{2\pi\cdot 10.2\Omega\cdot 624\textrm{MHz}}=25\textrm{pF}$$
Similar for $$\C_{jc}\$$. The pleasure of finding out the capacitance for 2N3904 belongs to you. BTW, maybe you know, pressing Ctrl+C when a transistor model is selected in the list, will copy the .model definition to the clipboard. That's how it's pasted in the schematic.
Using this method it's easy to determine $$\R_b\$$, $$\R_c\$$, and $$\R_e\$$, individually, by simply reading the flat portion of the voltage from the floating pin (V(y)). Since it's a current source of $$\1\textrm{A}\$$, the voltage at the floating pin will be $$\R_e\$$, in volts (for grounded emitter), and the difference will be $$\R_b\$$ (V(x,y)). Similar for $$\R_c\$$. This works better in .AC than .TRAN: